Timeline for Viscosity solutions of $(-\Delta)^s u = 0$ in $\Omega $ with non-homogeneous data $u = 1$ in $\mathbb R^n \setminus \Omega$
Current License: CC BY-SA 4.0
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| S Nov 12, 2021 at 12:03 | history | bounty ended | CommunityBot | ||
| S Nov 12, 2021 at 12:03 | history | notice removed | CommunityBot | ||
| S Nov 4, 2021 at 10:50 | history | bounty started | Zac | ||
| S Nov 4, 2021 at 10:50 | history | notice added | Zac | Improve details | |
| Nov 4, 2021 at 10:27 | comment | added | Mateusz Kwaśnicki | @Zac: Too long for a comment, posted as an answer. | |
| Nov 4, 2021 at 10:26 | answer | added | Mateusz Kwaśnicki | timeline score: 2 | |
| Nov 4, 2021 at 10:06 | history | edited | Zac | CC BY-SA 4.0 |
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| Nov 4, 2021 at 10:03 | comment | added | Zac | @MateuszKwaśnicki Yes, but the final result is surprising to me: I don't see how to show $$\int_\Omega \left( \int_0^\infty \lambda e^{-\lambda t} p_t^\Omega(x,y) dt \right) f(y) dy = 1 - \int_\Omega \left( \int_0^\infty \lambda e^{-\lambda t} p_t^\Omega(x,y) dt\right) dy$$ directly | |
| Nov 4, 2021 at 9:49 | comment | added | Mateusz Kwaśnicki | @Zac: This is called the Ikeda–Watanabe formula in the probability literature: the joint distribution of $(\tau, X_{\tau-}, X_\tau)$ is given by $p^\Omega_t(x, y) \nu(y,z) dt dy dz$, where $\nu(y,z) = c_{n,s} |y-z|^{-n-2s}$. From the PDE point of view, technical details aside, this is fairly straightforward, and you essentially described the argument in your question, did you not? | |
| Nov 4, 2021 at 8:58 | comment | added | Zac | @MateuszKwaśnicki Thank you so much. My confusion about (2) comes from this: formally, combining the approach in the OP and your comment, we should get $$u(x) = \int_\Omega \left(\int_{0}^\infty \lambda e^{-\lambda t} p_t^\Omega(x,y) dt\right) f(y) dy,$$ where $f(y) = c_{n,s} \int_{\mathbb R^n \setminus \Omega} |y - z|^{-n-2s} dz $. How can this shown to be the same as $u(x) = 1- \int_0^\infty\int_\Omega \lambda e^{-\lambda t}p_t^\Omega(x,y) dydt$? | |
| Nov 4, 2021 at 8:56 | comment | added | Mateusz Kwaśnicki | (...) It is quite straightforward to chech that $u(x)=\mathbb E^xe^{-\lambda \tau}$ satisfies $Lu=-\lambda u$, where $L$ is the Dynkin's characteristic operator. Boundary continuity of this $u$ follows from general facts about solutions of the Dirichlet problem. It remains to ask whether the Dynkin's characteristic operator is the same as $-(-\Delta)^s$. The answer is affirmative, but this is not entirely straightforward; for a partial discussion the relation between these two operators you may have a look at my survey at doi.org/10.1515/fca-2017-0002 | |
| Nov 4, 2021 at 8:52 | comment | added | Mateusz Kwaśnicki | @Zac: Regarding (2), unless $\Omega=\mathbb R^n$, we have $\int_\Omega p^\Omega_t(x,y)dy < 1$ both in the classical case ($s=1$) and in the fractional case ($0<s<1$). I am not aware of a simpler way to define the $\lambda$-Green function, even in the classical case. For (3), this is a general fact in the theory of semigroups of operators, any reference on that subject will do. An answer to (1) is more involved: this is a potential-theoretic fact, and in potential theory one usually tends not to use the generator whenever possible (except perhaps when discussing local operators). (...) | |
| Nov 4, 2021 at 8:28 | comment | added | Zac | @MateuszKwaśnicki I forgot to tag you in the follow-up questions in the comment above | |
| Nov 4, 2021 at 8:15 | comment | added | Zac | Thank you! A couple of follow up questions: (1) Why is $u(x) = \mathbb{E}^x e^{-\lambda \tau}$? (I'm quite ignorant about probability theory--do you have a reference for this?). (2) Is it possible to rewrite it more compactly in analytic terms? I was thinking something like using $\int_\Omega p_t^{\Omega} dx = 1$ (is this true in the fractional case too?) (3) Can you give me a reference about the fact that the integral is the Green function of $(-\Delta)^s + \lambda$?. | |
| Nov 3, 2021 at 18:53 | comment | added | Mateusz Kwaśnicki | @Zac: The integral $\int_0^\infty e^{-\lambda t} p_t^\Omega(x,y)dt$ is precisely the Green function for $(-\Delta)^s + \lambda$, or the $\lambda$-Green function for $(-\Delta)^s$. A "more compact way" to write this is, for example, $u(x) = \mathbb E^x e^{-\lambda \tau}$, where $\tau$ is the hitting time of $\Omega^c$ for the isotropic $2s$-stable Lévy process. :-) | |
| Nov 3, 2021 at 17:42 | comment | added | Zac | @MateuszKwaśnicki Thanks. Then, for $u$ we have $u(x) = 1- \int_0^\infty \int_\Omega \lambda e^{-\lambda t} p_t^{\Omega}(x,y) dy dt$. Can we write this in a more compact way? Also, why the heat kernel and not the Green function of the fractional Laplacian? | |
| Nov 3, 2021 at 9:55 | comment | added | Mateusz Kwaśnicki | Why not simply consider $v = 1-u$, which solves $(-\Delta)^s v + \lambda v = \lambda$ in $\Omega$, with homogeneous Dirichlet condition $v = 0$ in $\Omega^c$? The unique solution of the latter is given by $$v(x) = \int_0^\infty \int_\Omega \lambda e^{-\lambda t} p_t^\Omega(x, y) dy dt,$$ where $p_t^\Omega(x,y)$ is the corresponding heat kernel. | |
| Nov 3, 2021 at 9:48 | comment | added | Zac | @JackT Thank you! I've added a lower order term (which was the original model I had in mind) | |
| Nov 3, 2021 at 9:48 | history | edited | Zac | CC BY-SA 4.0 |
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| Nov 3, 2021 at 4:13 | comment | added | JackT | What is your motivation for this? Are you interested in viscosity solutions to equations involving the fractional Laplacian in general? I ask because $u \equiv 1$ is the (unique) strong solution | |
| Nov 2, 2021 at 6:48 | history | edited | Zac |
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| Nov 1, 2021 at 19:12 | history | edited | Zac | CC BY-SA 4.0 |
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| Nov 1, 2021 at 17:05 | history | edited | Zac | CC BY-SA 4.0 |
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| Nov 1, 2021 at 13:10 | history | asked | Zac | CC BY-SA 4.0 |